What is the Kinetic Theory of Gases, and how is it defined?

The kinetic energy of a gas refers to the capacity of each of its particles, which depends on the speed and, therefore, on the temperature to which it is subjected. Based on this concept, the diffusion of a gas allows it to move through a medium.

Both concepts, kinetic energy and diffusion in gases, are addressed by the Molecular Kinetic Theory which was developed by two scientists (Boltzmann and Maxwell) and explains the behavior of gases in general.

The function and variables in kinetic energy

In principle, the Theory describes variables such as the speed and kinetic energy of the particles and It relates them directly to other variables such as the pressure and temperature at which the gas is submit. Based on this, it is possible to describe that:

\(P = \;\frac{{m\; \cdot \;{v^2} \cdot \;N}}{{3 \cdot V}}\)

That is, the Pressure and Volume are related to variables of the molecule (m and N).

Based on the above, Maxwell and Bolzmann propose a mathematical function that can describe the distribution of the speeds of a gas as a function of its molar mass and temperature. It should be noted that this result is arrived at from a statistical analysis, where all the gas particles do not have the same speed, each one has its own speed, and from the distribution in the curve it is possible to find the speed value half. Finally, the average speed of a gas is said to be:

\(v = \sqrt {\frac{{3\;R\;T}}{M}} \)

Where the speed depends on the absolute temperature (T), the molar mass (M) and the universal gas constant (R).

Then, it can be interpreted that if different gases are at the same temperature, the one with the greater molar mass will have the lower average speed and vice versa. Likewise, if the same gas is exposed to two different temperatures, the one where the temperature is higher will have a higher average velocity, as is to be expected.

The concept of speed is closely related to the kinetic energy of the gas since:

\(Ec = \frac{1}{2}m{v^2}\)

The energy of a particle is a function of its average velocity. Now, for the gas, according to the Molecular Kinetic Theory it is known that the average value is given by:

\(\overline {Ec} = \;\frac{{3\;R\;T}}{2}\)

And it depends exclusively on the temperature.

diffusion in gases

When we talk about gases, to define them, we can mention different properties. For example, we can talk about its density, its viscosity, its vapor pressure as well as many other variables. One of them (and a very important one) is dissemination.

Diffusion is related to the ability of the same to move in a certain medium. In general, diffusion is related to the "driving forces" that allow fluid migration from one side to another. For example, the diffusion of the gas depends on many parameters, such as whether there is a pressure difference between points A and B towards which it moves, or a difference in concentrations. In turn, it also depends on factors such as the temperature and the molar mass of the gas, as seen above.

Based on the above, Graham studied the behavior of gases in terms of their diffusion and emulated a Law that establishes that:

"At constant pressure and temperature, the diffusion rates of different gases are inversely proportional to the square root of their densities." In mathematical terms it is expressed as follows:

\(\frac{{{v_1}}}{{{v_2}}} = \;\sqrt {\frac{{{\rho _2}}}{{{\rho _1}}}} \)

Being v1 and v2 the speeds of the gases and \(\rho \) their densities.

If we work mathematically with the previous expression we get:

\(\frac{{{v_1}}}{{{v_2}}} = \;\sqrt {\frac{{{M_2}}}{{{M_1}}}} \)

Since M1 and M2 are the molar masses respectively and, if the pressure and temperature do not vary, the relationship between them is identical to the relationship between the densities of gases.

Finally, Graham's Law expresses the above in terms of diffusion time. If we consider that both gases must diffuse along the same length and at the speed v1 and v2 previously determined, it can be said that:

\(\frac{{{t_1}}}{{{t_2}}} = \;\sqrt {\frac{{{M_2}}}{{{M_1}}}} \)

Finally, we can deduce that a gas with a higher molar mass will have a longer diffusion time than a gas with a lower molar mass, if both are subjected to the same conditions of temperature and pressure.